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Mumz [18]
3 years ago
11

The combustion of glucose, C6 H12 O6 (s), produces carbon dioxide, CO2 (g), and water, H2 O(g), according to the equation below.

mc020-1.jpg The enthalpy of the reaction is –2,840 kJ. What is the heat of combustion, per mole, of glucose? –2,840 kJ/mol –473.3 kJ/mol 473.3 kJ/mol 2,840 kJ/mol
Chemistry
2 answers:
Mice21 [21]3 years ago
8 0

Answer:  –2,840 kJ/mol

Explanation: C_6H_{12}O_6+O_2\rightarrow 6CO_2+6H_2O  \Delta H=-2840kJ

Heat of combustion is the heat produced when 1 mole of the substance is completely oxidized. The production of heat means the reaction is an exothermic process and gives off energy.

According to the given balanced chemical equation:

1 mole of glucose is being completely oxidized to give carbon dioxide and water and thus heat of combustion is -2840 kJ/mol.

Hoochie [10]3 years ago
3 0

Answer:

the answer is A

Explanation:

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                                 C₂H₆  +  O₂    →    CO₂  +  H₂<span>O

There are 2 C at left hand side and 1 carbon at right hand side. So, multiply CO</span>₂ by 2 to balance C atoms at both side. So,

                                 C₂H₆  +  O₂    →   2 CO₂  +  H₂O

Now, count number of H atoms at both sides. There are 6 H atoms at left hand side and 2 at right hand side. Multiply H₂O by 3 to balance H atoms.


                                 C₂H₆  +  O₂    →   2 CO₂  +  3 H₂O

At last, balance O atoms. There are 2 O atoms at left hand side and 3 O atoms at right hand side. Multiply O₂ with 1.5 (i.e. 3/2) to balance O atoms. i.e.

                                 C₂H₆  +  3/2 O₂    →   2 CO₂  +  3 H₂O

Hence, the equation is balanced. If you want to make equation fraction free then multiply all equation with 2. i.e.

                           ( C₂H₆  +  3/2 O₂    →   2 CO₂  +  3 H₂O ) × 2

                                2 C₂H₆  +  3 O₂    →   4 CO₂  +  6 H₂O
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